Two point charges 4Q, Q are separated by 1 m in air. At what point on the line joining the charges is the electric field intensity zero?
Also calculate the electrostatic potential energy of the system of charges, taking the value of charge, Q = 2 x 10^–7 C.

Solution

Given, two point charges 4Q and Q are kept 1m apart.
To find:
(i) The point where, electric field intensity (E) is 0 on the line joining the charges.
(ii) Electrostatic potential energy (U) of the system of charges.

Let, the required point be P, at a distance x from charge 4Q and, at (1-x) distance from charge Q.

(i) Electric field at P due to charge 4Q = Electric field at P due to Q

E = \frac{k q}{r}

Therefore, electric field intensity is 0 at x=

\frac{2}{3} m

from charge 4Q.

(ii) Electrostatic potential energy of the system is given as,

Therefore, electrostatic potential energy of the ststem of charges is equal to

1.44 \times 10^{- 3} J

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Electrostatic Potential And Capacitance

Two point charges 4Q, Q are separated by 1 m in air. At what point on the line joining the charges is the electric field intensity zero?
Also calculate the electrostatic potential energy of the system of charges, taking the value of charge, Q = 2 x 10^–7 C.

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Electrostatic Potential And Capacitance

Two point charges 4Q, Q are separated by 1 m in air. At what point on the line joining the charges is the electric field intensity zero?Also calculate the electrostatic potential energy of the system of charges, taking the value of charge, Q = 2 x 10–7 C.

Some More Questions From Electrostatic Potential and Capacitance Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Two point charges 4Q, Q are separated by 1 m in air. At what point on the line joining the charges is the electric field intensity zero?
Also calculate the electrostatic potential energy of the system of charges, taking the value of charge, Q = 2 x 10^–7 C.